mersenneforum.org Factors of Fibonacci numbers
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 2020-10-09, 06:52 #1 RMLabrador     "Roman V. Makarchuk" Aug 2020 Ukraine 2·17 Posts Factors of Fibonacci numbers Greetings. The factors of Fibonacci numbers: For prime p: $F(p)=\prod_{i=1}^{m} (k_{i}\cdot p\pm 1)$ For an any n $n=p_{1}^{t1}\cdot p_{2}^{t2}\cdot...\cdot p_{k}^{tk}$ $F(n)=\prod_{i=1}^{m} (k_{i}\cdot p_{1}\pm 1)\cdot (l_{i}\cdot p_{2}\pm 1)\cdot ...\cdot(z_{i}\cdot p_{k}\pm 1)$ Example p=103 F(103)=1500520536206896083277=519121*5644193*512119709 (519121-1)/103=5040 (5644193+1)/103=54798 (512119709-1)/103=4972036 P.S. Of course,m and coefficients k both are function of p or n)
 2020-10-10, 08:31 #2 RMLabrador     "Roman V. Makarchuk" Aug 2020 Ukraine 2·17 Posts Interesting. It is not so hard to write some simple code, like this (Pari GP) forprime(n=11,900,x=Mod(factorint(fibonacci(n)),n);print(n,lift(x))) and see, that posted above is true for all prime, at least for those, that in the scope of your patience)) Taking into account the level of your obsession with factorization 2 ^ n + 1 or so, my next post will be about it)))
2020-10-11, 00:27   #3
Dr Sardonicus

Feb 2017
Nowhere

43·97 Posts

Quote:
 Originally Posted by RMLabrador Greetings. The factors of Fibonacci numbers: For prime p: $F(p)=\prod_{i=1}^{m} (k_{i}\cdot p\pm 1)$ For an any n $n=p_{1}^{t1}\cdot p_{2}^{t2}\cdot...\cdot p_{k}^{tk}$
I note that F5 = 5 does not conform to your prescription.

The following is well known. It is an exercise for beginners in the subject of Fibonacci and Lucas numbers.

Let p be a prime number.

1) If (5/p) = +1 [p == 1 or 4 (mod 5)], then p divides Fp-1. Corollary: If (5/p) = +1, and p divides Fq with q prime, then q divides p-1, i.e. p == 1 (mod q).

2) If (5/p) = -1 [p == 2 or 3 (mod 5)], then p divides Fp+1. Corollary: If (5/p) = -1, and p divides Fq with q prime, then q divides p + 1, i.e. p == -1 (mod q).

 2020-10-11, 16:30 #4 RMLabrador     "Roman V. Makarchuk" Aug 2020 Ukraine 428 Posts i guess, "well known" is not true)) here, no mention about http://www.maths.surrey.ac.uk/hosted....html#section2 and i can find 10+ links like this, and 0 link with similar formulae, like mine. I just state the global form of factors of Fibonacci numbers, and not properties of F(n) itself. Besides, 2,3,5 and 7 in my opinion are not prime. They uber prime, they posses the rule of 1 and themselves division, and the broke the many other, i.e. 2 is even, 3 and 5 and 7 are too monadic) remember 6k+/-1? 6=2*3 and 6=5+1 and 6=7-1. Forget this for a while.
 2020-10-11, 16:58 #5 RMLabrador     "Roman V. Makarchuk" Aug 2020 Ukraine 2·17 Posts Just run this in some other than Pari/GP system x = 251; c = Fibonacci[x]; (*Solve[(((a*x+c-1)/((a*x-1)*x))^(-1))-(1/b)==0&&(x*a-1)*(b*x-1) ==c,{a,b},Integers]*) Solve[((a + b)^2 + 4*a*b*(c - 1) - z^2 == 0) && (x*a - 1)*(b*x - 1) - c == 0 && b > a && a > 0 && z > 0, {a, b, z}, Integers] Last fiddled with by RMLabrador on 2020-10-11 at 17:03 Reason: clear mind

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